Evaluation of Inter-strand Contact Force

This section takes a discrete analysis approach in examining the contacts within the conductor with a view of identifying the various forces acting. For a single layer conductor as shown in figure (2.12), the contact between the only layer and the core are line contact and the stresses are analysed along the line contact as a form of pressure distribution, shown in figure (2.13). For a multilayer conductor as shown in figure (2.14), while, apart from the contact between the first layer and the core, other forms of contact between layers are actually point contacts. It is well known that most stresses are located in the area of point contacts. Though, most inter-strand contacts are point contacts, in most studies, a line contact of an equivalent point is normally used. Similarly, also used here, concepts associated with the already well-established line contact stresses were used to deduce the various forces that acts within the conductor [13, 52, 57].

Presented as follows is the combined summarized derivations, as documented in [13, 52, 57]. For the derivation, a single layer conductor as shown in figure (2.12) was considered first and later extended to the multilayer conductor.

Figure 2.12: Single-layered conductor

To evaluate the various internal forces acting within the conductor, the analysis starts when the conductor is assumed to be in the stick condition and only the axial force acting on each strand is considered. Under the stick condition there is no relative displacement between strands. This is made possible because the friction force at the contacting regions is greater than the shearing forces tending to pull the strands apart. It is assumed that both strands exert equal and opposite force at these regions of contacts. Also, the only internal loads acting on a conductor cross-section is assumed to be the normal force (tension) and the small bending moment arising from the strand curvature.

37

Figure 2.13: The normal and frictional forces in a single-layered conductor In line with the above, considering the outer layer forces of the single layer conductor (i1 or N

= 1) as shown in figure (2.13), with the conductor cross-section subjected to axial loads, S due to tensioning acting at both ends. The axial load induces tension on individual strands _Ti and this tension has two components due to the helical arrangement: axial T_i,Aand tangential T_i,T components. Neglecting the tangential components, the tensioning force on each individual strand in given layer of the conductor is therefore calculated as:

i A i i

T T

 cos

 ,

.…..………….……. (2.27) Under this static condition with the presence of the conductor’s axial tension S, each individual strand can be assumed to be subject to the same tensile force as given by [57]

EA S A T E

i i i

i

i  



^3

2

cos

cos .……….……. (2.28) The axial load S, acting on the conductor cross-section tends to straighten the strands, thereby inducing pressure on the core below. For the single layer conductor shown in figure (2.13), the layer exerts a pressure on the core and this can be evaluated as:

 







 _{i S}

S

i T

d d

p 2T^, sin 2 ^,





……… (2.29) where ρ is the radius of curvature and β is the wrap angle

Considering a curve differential element of the strand of an arc length, given as



 

 sin

d rd

ds  . .……….……. (2.30) The exerted pressure gives rise to the radial force along the arc length of the differential element.

Hence, this radial force is calculated as







ds T d

F_N    _i,Ssin _i .……….……. (2.31)

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The surface along the line contact gives rise to the component of normal force. Mathematically, at the surface for a two contacting strands:

 i N

S F

dF 



.……….……. (2.32) Substituting equations (2.30) and (2.31) into equation (2.32), the radial force that acts on the individual strand under tension is obtained as:







T d

dF_S(i)  _i,Ssin _i .……….……. (2.33) If the frictional effects is said to be greater than the shear force that is tending to pull the strands apart and assuming the sliding is just about to occur, which implies that friction force is maximum.

The reaction force dTi, S on an individual strand due to friction can be calculated as:

 



dF



T



d



dF

dT_i,S  _{S i}  _N  _i,Ssin _i .……….……. (2.34) The summation of all the normal forces for the entire cross sections is obtained by integrating the above first order differential equation and the solution is given as:



 ^sin

, ( ) Ce

T_{i S}  _i .……….……. (2.35) The constant Ci is evaluated with the above equation satisfying the boundary condition: where



0 and T_i,S(0)C_ie^sin . Thus EA S

A T E

C

i i i

i i

i .

cos cos

3 2





 

 .…..………….……. (2.36)

Therefore, the normal force acting, is given as: T_i,S(0)T_i

In the no-slip condition, the force exerted due to bending can be calculated according to the usual Bernoulli-Euler beam bending theory. Hence, the total force Ti, S for the stick state is calculated and this will yield the function:







T e ⁱ

T_stick( ) _i ^sin .……….……. (2.37) The above equation gives the value for the normal force that acts just before bending i.e. when the strand is just about to slip. During bending under dynamic loading, the frictional force is overcome resulting in relative sliding between the strands in point contact. Sliding occurs when reaction force is greater than friction force assuming that the tensile force is constant along the strand. Thus, beyond a given bending amplitude, micro-slip occurs at the inter-layer contact points. During slippage, the normal force is reduced and this can be calculated as

39 )

1 (

) ( )

( T  T T e^{sini} 

T_{slip stick i i} .……….……. (2.38)

Figure 2.14: Double-layered conductor

Next, was to extend the above procedure used to determine the normal force acting on a single layer to a multilayer conductor. Starting with double layer conductor shown in figure (2.14). In analysing a multilayer conductor, the above derivation done for the single layer above the core is applicable to the outer most layer of the multilayer conductor.

Figure 2.15: The normal and frictional forces in a double-layered conductor

Considering each layer of a multilayer conductor, shown in figure (2.14) which is a doubled- layered conductor subjected to axial tension, S. Firstly, consider the double layer conductor shown in figure (2.15), the number of layer i =1, 2 or N=2. In this case, extending already derived equations for a single layer conductor as applicable to the outer layer where N = 2.

Now considering the derivation for the penultimate layer, i =1 or N-1 i.e. layer just below the outer layer. The normal forces acting on the outer part of the strand with the outer strand can be determined as:

_i,_outer _N,_N1 _N1sin _{i N}1

N dF T d

dF

 

.……….……. (2.39) Also, the normal forces acting on inner part of the strand with the inner strand can be determined as:

40

1 1 1

2 , 1 )

.

(_iinner  _{N N}  _N sin _{N N}

N dF T d

dF

 

.……….……. (2.40) Therefore, the imbalance force acting on the strands in this layer is by the summation of equations (2.39) and (2.40), then multiply by its coefficient of frictions.

) ( _{, 1 (, ) 1, 2 (, )}

1 N N N iouter N N N iinner

N dF dF

dT _ 



_ 



_{ } .……….……. (2.41) The substitution of equations (2.39 and 2.40) into equation (2.41) result to

 

_{1 1 1 1}



₁



₁

1 _{ }

sin

_{  }

sin

_





_{N N N N}



_N



_{N N N N}

N

T d T d

T

d        

.…..………….……. (2.42)

The solution to the above differential equation comprise of the homogeneous and the particular solutions given as

 _{N N} _{N N}

N N

N T e

e T

T_N1(



_N1)  _N1 ^{ 1 1sin 1}  _N ^{ 1  1sin}

.……….……. (2.43) The evaluation of the above equation is by satisfying the boundary conditions as the same for single layer conductor, but in this case the constants T_N1and T_N must be determined. Because in this study, the conductors used for analysis is ACSR, thus a constant co-efficient of friction is not assumed. Applying similar boundary condition for the multilayer conductor:

Where ^{1 1 1} ,

 

^{ 1 sin 1}

sin 1 ,

1 (0) 0

,

0 _  _ ^{  }  ^{ }

 T_{N S} T_N e^{N N N} and T_NS C_Ne^{N N NN}

 .

The normal force for the stick condition is calculated by

 



N N

 ^

N N

^

stick T e T e T T

T (



) _1 ^{N1sinN1N1} 2 ^{N1N sinNN1}  _1 .……….……. (2.44) For the slip condition, this force becomes an exponential function whose exact form depends on the selected slip state. Taking this into account, the force is given by

 



²

 ^{ }

¹

)

(  _N1 ^{1sin 1 1}  _N ^{1 sin 1}  _N1 _N 

slip T e T e T T

T



^{N N N N N NN} …….……. (2.45)

The above derivations can now be extended to the case of various normal forces between each strand, for any given layer for the general case of the conductor geometry, with an arbitrary number of layers in the conductor.

Generally, the normal force acting in the stick condition for any arbitrary number of multilayer conductor can be calculated by

 

  



  

 

  



 



 



^{  N}

i

i i N

i i i

stick

T T e T T

T

^{i i i i}

1

1 sin

1 1

1

2

1

)

( 

^{   } .……….……. (2.46)

For the slip condition

 

 ¹ 

2 )

(

^{1 sin 1}

1

1

 



 



 



^{ }



 ^T



^e

^{i i i i}

T T

N

i i i

slip









 .…..………….……. (2.47)

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2.9 Characterization of the Conductor Cross-Sectional Parameters